sallyhensenOPI
sallyhensenOPI sallyhensenOPI
  • 17-04-2019
  • Chemistry
contestada

What is the Ka of a 0.653 m solution of hydrocyanic acid with a ph of 5.47

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superman1987
superman1987 superman1987
  • 17-04-2019

Answer:

1.758 x 10⁻¹¹.

Explanation:

∵ pH = - log[H⁺].

∴ 5.47 = -log[H⁺].

log[H⁺] = -5.47.

∴ [H⁺] = 3.388 x 10⁻⁶ M.

∵ [H⁺] = √(Ka.C)

∴ [H⁺]² = Ka.C

∴ Ka = [H⁺]²/C = (3.388 x 10⁻⁶)²(0.653) = 1.758 x 10⁻¹¹.

Answer Link
bartolosaul6
bartolosaul6 bartolosaul6
  • 12-12-2021

Answer:

The answer is actually 1.76x10^-11

Explanation:

Fill in the box 1.76 and then -11

Answer Link

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